Question

EXAM
A vehicle of mass 1250 kg is driven
ko is driven with an
on 0.2 ms' along a straightley
acceleration 0.2 ms al
al resistive force 500 N
delivered by the
road against an external resistive for
Calculate the power delivered b
vehicle's engine if the velocity of the ye
he if the velocity of the vehice
Ade
is 30 ms. slove this sum​

Answer 1

Explanation:

Answer:

\fbox {a = - 1.5 m/s^2}

a=−1.5m/s

2

\fbox {Distance (s) = 300m}

Distance(s)=300m

===========================

Given data :-

Initial velocity (u) = 30m/s

Final velocity (v) = 0

Since the body comes to rest after applying breaks , so final velocity is " zero "

Time taken :- 20 s

=============================

Question :- To find deceleration and distance covered

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FOR Solving Problem Like THESE We Use 3 Equations Of Uniform Acceleration .They Are

(1) V = u + at

(2) v^2 - u^2 = 2as

(3) S = ut + 1/2 at^2

We use equation (1) to find deceleration because the rest 2 equations has " s " in it ( s is unknown value )

\fbox {V = u + at}

V=u+at

0 = 30 + a (20)

30 + 20a =0

20 a = -30

a = -30/20

a = - 3/2

a = -1.5 m/s^2

Now it's time to calculate " distance"

\fbox {V^2 - u^2 = 2as}

V

2

−u

2

=2as

0 - (30)^2 = 2 (-1.5)s

- 900 = -3S

900 = 3s

S = 900/3

S = 300m

◇ THING TO KNOW :-

Acceleration in negative direction is called as deceleration.